Using the Quadratic Formula - Math

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A baseball that is thrown in the air follows a trajectory of , where is the height of the ball in feet and is the time elapsed in seconds. How long does the ball stay in the air before it hits the ground?

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To solve this, we look at the equation .

Setting the equation equal to 0 we get .

Once in this form, we can use the Quadratic Formula to solve for .

The quadratic formula says that if , then

$x= $$\frac{-b\pm\sqrt(b^2$-4ac)}{2a}$$ .

Plugging in our values:

$t= $$\frac{-12\pm\sqrt{(12^2$-4(-4)(6)}$}{2(-4)}=\frac{-12\pm \sqrt{240}$}{-8}= $\frac{12\pm15.5}{8}$$

Therefore or and since we are looking only for positive values (because we can't have negative time), 3.4375 seconds is our answer.

Solve using the quadratic formula:

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Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(2) \pm $\sqrt{(2)^2$-4(2)(3)}$}{2(2)}$

$x = $\frac{-2 \pm \sqrt{-20}$}{4}$

$x = $\frac{-2 \pm 2i\sqrt{5}$}{4}$

$x = $\frac{-1 \pm i\sqrt{5}$}{2}$

Solve using the quadratic formula:

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Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$a = $\frac{-(-5) \pm $\sqrt{(-5)^2$-4(2)(4)}$}{2(2)}$

$a = $\frac{5 \pm \sqrt{-7}$}{4}$

$a = $\frac{5 \pm i\sqrt{7}$}{4}$

Solve using the quadratic formula:

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Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(b) \pm $\sqrt{(b)^2$-4(a)(3b)}$}{2(a)}$

$x = $\frac{-b \pm $\sqrt{b^2$-12ab}$}{2a}$

Solve using the quadratric formula:

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Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(ac) \pm $\sqrt{(ac)^2$-4(b)(c)}$}{2(b)}$

$x = $\frac{-ac \pm $\sqrt{a^2c$^2-4bc}$}{2b}$

A baseball that is thrown in the air follows a trajectory of , where is the height of the ball in feet and is the time elapsed in seconds. How long does the ball stay in the air before it hits the ground?

Tap to see back →

To solve this, we look at the equation .

Setting the equation equal to 0 we get .

Once in this form, we can use the Quadratic Formula to solve for .

The quadratic formula says that if , then

$x= $$\frac{-b\pm\sqrt(b^2$-4ac)}{2a}$$ .

Plugging in our values:

$t= $$\frac{-12\pm\sqrt{(12^2$-4(-4)(6)}$}{2(-4)}=\frac{-12\pm \sqrt{240}$}{-8}= $\frac{12\pm15.5}{8}$$

Therefore or and since we are looking only for positive values (because we can't have negative time), 3.4375 seconds is our answer.

Solve using the quadratic formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(2) \pm $\sqrt{(2)^2$-4(2)(3)}$}{2(2)}$

$x = $\frac{-2 \pm \sqrt{-20}$}{4}$

$x = $\frac{-2 \pm 2i\sqrt{5}$}{4}$

$x = $\frac{-1 \pm i\sqrt{5}$}{2}$

Solve using the quadratic formula:

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Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$a = $\frac{-(-5) \pm $\sqrt{(-5)^2$-4(2)(4)}$}{2(2)}$

$a = $\frac{5 \pm \sqrt{-7}$}{4}$

$a = $\frac{5 \pm i\sqrt{7}$}{4}$

Solve using the quadratic formula:

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Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(b) \pm $\sqrt{(b)^2$-4(a)(3b)}$}{2(a)}$

$x = $\frac{-b \pm $\sqrt{b^2$-12ab}$}{2a}$

Solve using the quadratric formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(ac) \pm $\sqrt{(ac)^2$-4(b)(c)}$}{2(b)}$

$x = $\frac{-ac \pm $\sqrt{a^2c$^2-4bc}$}{2b}$

A baseball that is thrown in the air follows a trajectory of , where is the height of the ball in feet and is the time elapsed in seconds. How long does the ball stay in the air before it hits the ground?

Tap to see back →

To solve this, we look at the equation .

Setting the equation equal to 0 we get .

Once in this form, we can use the Quadratic Formula to solve for .

The quadratic formula says that if , then

$x= $$\frac{-b\pm\sqrt(b^2$-4ac)}{2a}$$ .

Plugging in our values:

$t= $$\frac{-12\pm\sqrt{(12^2$-4(-4)(6)}$}{2(-4)}=\frac{-12\pm \sqrt{240}$}{-8}= $\frac{12\pm15.5}{8}$$

Therefore or and since we are looking only for positive values (because we can't have negative time), 3.4375 seconds is our answer.

Solve using the quadratic formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(2) \pm $\sqrt{(2)^2$-4(2)(3)}$}{2(2)}$

$x = $\frac{-2 \pm \sqrt{-20}$}{4}$

$x = $\frac{-2 \pm 2i\sqrt{5}$}{4}$

$x = $\frac{-1 \pm i\sqrt{5}$}{2}$

Solve using the quadratic formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$a = $\frac{-(-5) \pm $\sqrt{(-5)^2$-4(2)(4)}$}{2(2)}$

$a = $\frac{5 \pm \sqrt{-7}$}{4}$

$a = $\frac{5 \pm i\sqrt{7}$}{4}$

Solve using the quadratic formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(b) \pm $\sqrt{(b)^2$-4(a)(3b)}$}{2(a)}$

$x = $\frac{-b \pm $\sqrt{b^2$-12ab}$}{2a}$

Solve using the quadratric formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(ac) \pm $\sqrt{(ac)^2$-4(b)(c)}$}{2(b)}$

$x = $\frac{-ac \pm $\sqrt{a^2c$^2-4bc}$}{2b}$

A baseball that is thrown in the air follows a trajectory of , where is the height of the ball in feet and is the time elapsed in seconds. How long does the ball stay in the air before it hits the ground?

Tap to see back →

To solve this, we look at the equation .

Setting the equation equal to 0 we get .

Once in this form, we can use the Quadratic Formula to solve for .

The quadratic formula says that if , then

$x= $$\frac{-b\pm\sqrt(b^2$-4ac)}{2a}$$ .

Plugging in our values:

$t= $$\frac{-12\pm\sqrt{(12^2$-4(-4)(6)}$}{2(-4)}=\frac{-12\pm \sqrt{240}$}{-8}= $\frac{12\pm15.5}{8}$$

Therefore or and since we are looking only for positive values (because we can't have negative time), 3.4375 seconds is our answer.

Solve using the quadratic formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(2) \pm $\sqrt{(2)^2$-4(2)(3)}$}{2(2)}$

$x = $\frac{-2 \pm \sqrt{-20}$}{4}$

$x = $\frac{-2 \pm 2i\sqrt{5}$}{4}$

$x = $\frac{-1 \pm i\sqrt{5}$}{2}$

Solve using the quadratic formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$a = $\frac{-(-5) \pm $\sqrt{(-5)^2$-4(2)(4)}$}{2(2)}$

$a = $\frac{5 \pm \sqrt{-7}$}{4}$

$a = $\frac{5 \pm i\sqrt{7}$}{4}$

Solve using the quadratic formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(b) \pm $\sqrt{(b)^2$-4(a)(3b)}$}{2(a)}$

$x = $\frac{-b \pm $\sqrt{b^2$-12ab}$}{2a}$

Solve using the quadratric formula:

Tap to see back →

Use the quadratic formula to solve:

$x = $\frac{-b \pm $\sqrt{b^2$-4ac}$}{2a}$

$x = $\frac{-(ac) \pm $\sqrt{(ac)^2$-4(b)(c)}$}{2(b)}$

$x = $\frac{-ac \pm $\sqrt{a^2c$^2-4bc}$}{2b}$